Jupiter is 5.2 times farther from the Sun than Earth is. Find Jupiters orbital period in Earth years...?

Question

Please help me with this problem.
It seems so easy.
But I'm having trouble setting it up.

Answer 1

Ta^2 / Tb^2 = Ra^3 / Rb^3

T=period
R=orbital radius

Ta (earth) = 1 year
Tb (jupiter) = ?
Ra (earth) = 1 Au (astronomical unit)
Rb (jupiter) = 5.2 AU


1^2 / Tb^2 = 1^3 / 5.2^3

1 x 5.2^3 = Tb^2

Tb = (5.2^3) ^ .5

Tb = 11.86 years

Answer 2

Ian Ridpath in ASTRONOMY, published by DK Publishing, NY, NY, states that the orbital period of Jupiter around the Sun is 11.86 Earth Years.

The problem you face in trying to calculate the orbital period is that you are missing "velocity of travel."

Jupiter is 483.7 million miles from the Sun (on average).
Double that ( = 967.4 million miles ) and multiply it by 3.14 to get the length of a typical circle orbit (Jupiter's orbit is not circular).
My math shows that to be 3037.6 Million Miles orbit length.

Now divide
3,037.6 million miles
by
103,893 hours

to get Jupiter's speed around the orbital path.

Discarding some significant digits I calculate that to be
2.92 x 10^4 miles per hour

or 29,200 miles per hour.

Unless I have made a mistake, that speed should get Jupiter around its orbit in 11.86 years or so. And, at 29,000 mph, I would hesitate to get in its path.

That produces a circumference of a circle we could suggest as being Jupiter's path of orbit. But given all those miles of orbit distance, few people ever mention how fast (in miles per hour) Jupiter is flying. In the math above I have tried to do that for you.

Regards,
Zah


So if you are into Math...
11.86 Earth Years x 365 days per year = 4,328.9 Days to make one orbit.

4,328.9 Days x 24 hours per day = 103,893.6 hours /orbit.

Answer 3

The semi-major axis can be assumed to be 5.2* bigger than the earth's. So the period will be 5.2^(3/2)*as long as the earth's.